To show that every quotient group of an abelian group is abelian, we need to show that the operation in the quotient group is commutative.

Let G be an abelian group and let N be a normal subgroup of G. We define the quotient group G/N as the set of cosets of N in G, with the operation defined as (aN)(bN) = (ab)N for all a, b in G.

To show that G/N is abelian, we need to show that (aN)(bN) = (bN)(aN) for all a, b in G.

Let x be any element in aN, which means that x = an for some n in N. Then (aN)(bN) = (an)(bN) = (ab)nN = (ba)N = (ba)(aN) = (bN)(aN), where the second equality follows from the commutativity of the operation in G, and the fourth equality follows from the fact that N is a normal subgroup.

Since (aN)(bN) = (bN)(aN) for all a, b in G, we conclude that every quotient group of an abelian group is abelian.

b)(ii) To show that every quotient group of a cyclic group is not cyclic, we need to find a specific example of a cyclic group and a normal subgroup such that the quotient group is not cyclic.

Consider the cyclic group G = Z (the additive group of integers) and let N = 2Z (the subgroup of even integers).

The quotient group G/N consists of all cosets of N in G, which are of the form {a + 2Z | a is an integer}.

To show that G/N is not cyclic, we need to show that there is no single element in G/N that generates the entire group.

Suppose there exists an element g + 2Z in G/N that generates the entire group. This means that for every element a + 2Z in G/N, there exists an integer k such that (g + 2Z)^k = a + 2Z.

Let's choose a = 1 + 2Z. Then we have (g + 2Z)^k = 1 + 2Z for some integer k.

Expanding the left side, we have g^k + 2Z = 1 + 2Z. This implies that g^k - 1 is an even integer, which means that g^k is an odd integer.

However, since g is an integer, g^k can be either even or odd depending on the value of k. Therefore, it is not possible for (g + 2Z)^k = 1 + 2Z for all integers k.

Therefore, there is no single element in G/N that generates the entire group, and we conclude that every quotient group of a cyclic group is not cyclic.